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B Thermodynamic equation from textbook

  1. Jan 31, 2016 #1
    In my text it was said that ΔH=ΔU+ΔnRT
    H:ENTHALPY
    U:INTERNAL ENERGY
    Δn:n of gaseous products-gaseous reactents
    they use ΔnRT instead of Δ(PV),here T is constant⇒U is a constant(U is a function of temperature)
    then why didn't the term ΔU became zero?
    what is the actual equation?is it is ΔH=ΔU+ΔnRT
    please help
     
  2. jcsd
  3. Jan 31, 2016 #2

    QuantumQuest

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    The equation is ΔU = ΔH – Δng RT or equivalently the way you wrote it. Now, what does this tell you about internal energy change when you have a gas (or not)?
     
    Last edited: Jan 31, 2016
  4. Jan 31, 2016 #3
    If a chemical reaction is involved, ΔH and ΔU are not just functions of temperature. They are also functions of the amounts of the various materials present, and these have changed. Have you learned about heats of formation or heats of reaction?
     
  5. Jan 31, 2016 #4

    QuantumQuest

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    I didn't say that they are. I just point out how the system interchanges energy with its surroundings and what does this imply for the various parameters of the system.
     
  6. Jan 31, 2016 #5
    I wasn't responding to you. I was responding to the OP.
     
  7. Feb 1, 2016 #6
    but how in a chemical reaction(in general) the amount of mass change(since U changes when 1.heat passes 2.work done 3.matter enters or leaves ) it will violate the law of mass conservation.
    i think if a reaction takes place at constant T the heat evolved(ΔH) must fully use for doing work(expansion or compression) then ΔU must be 0.
    please explain about dependence of ΔU to ΔT too
     
  8. Feb 1, 2016 #7
    If a chemical reaction takes place in a closed system, then mass is conserved.
    You may think this, but it is not correct. Have you considered the amount of energy involved in making and breaking chemical bonds? When chemical reactions occur, this affects ΔU and ΔH. If an exothermic reaction occurs at constant pressure and temperature, ΔH is negative, and heat must be removed from the reactor in order to hold the temperature constant. If the reaction were carried out adiabatically, the temperature would rise.

    The internal energy and enthalpy of a reaction mixture is a function not only of temperature, but also of the number of moles of the various chemical species present.

    $$U = U(T,P, n_1,n_2,...)$$
    $$H = H(T,P, n_1,n_2,...)$$

    So, at constant temperature and pressure, if the number of moles of the various species in the mixture changes (while conserving overall mass in a closed system), both U and H will change.
     
  9. Feb 2, 2016 #8
    You may think this, but it is not correct. Have you considered the amount of energy involved in making and breaking chemical bonds? When chemical reactions occur, this affects ΔU and ΔH. If an exothermic reaction occurs at constant pressure and temperature, ΔH is negative, and heat must be removed from the reactor in order to hold the temperature constant. If the reaction were carried out adiabatically, the temperature would rise.

    The internal energy and enthalpy of a reaction mixture is a function not only of temperature, but also of the number of moles of the various chemical species present.

    $$U = U(T,P, n_1,n_2,...)$$
    $$H = H(T,P, n_1,n_2,...)$$

    So, at constant temperature and pressure, if the number of moles of the various species in the mixture changes (while conserving overall mass in a closed system), both U and H will change.[/QUOTE]
    Then will the ΔU at constant volume for a combustion reaction equal to that of constant pressure(temperature constant)
    (will be helpful to find ΔH at constant P if ΔH at constant V and Δng are known)
    an extra question:what is the state of H2O in combustion of methane(liquid or gas)
    [/QUOTE]
     
  10. Feb 2, 2016 #9
    Suppose that, after you measure ΔU at constant volume, you cause the volume of your ideal gas mixture to increase (or decrease) by adding (or removing) heat while holding the temperature constant until you reach the original pressure. What would be the change in U for this second process? What would be the overall change in U?
    It depends on how much heat you remove. If you don't remove enough heat to condense the vapor at the temperature 25C, then it will be gas. If you remove enough heat to condense the vapor at the tempearture 25C, then it will be liquid.
     
  11. Feb 5, 2016 #10
    we take two separate systems and allow the same reaction to proceed in both with one at constant V and other at constant P(at constant T)
     
  12. Feb 5, 2016 #11
    You haven't answered my questions in post #9. The answer to those questions will help you answer this.
     
  13. Feb 5, 2016 #12
    in your post the T is constant so the supplied heat must be fully use to do work so the in U from constant V to that P will be zero
    so overall change will be ΔU
    But here the reaction takes place at constant V not at constant P.(we add or remove heat after the reaction)
     
  14. Feb 5, 2016 #13
    Multiple choice question: Internal energy is

    (a) a function of state

    (b) not a function of state?
     
  15. Feb 5, 2016 #14
    it is a function of state
     
  16. Feb 5, 2016 #15
    OK.

    State 1: Reactants at temperature T, pressure P, and volume V

    State 2: Products at temperature T, volume V, pressure ≠ P

    State 3: Products at temperature T, volume ≠ V, pressure P

    Change in internal energy from State 1 to State 2 = (ΔU)V

    Change in internal energy from State 2 to State 3 = 0

    Change in internal energy from State 1 to State 3 = (ΔU)P = (ΔU)V + 0 = ??
     
  17. Feb 7, 2016 #16
    yeah I get you.
    so for a isothermal reaction process ΔU is the energy used for bond dissociation,formation,etc.the excess heat will removed sometimes by doing a work.
    1.I heard phase transformation is a isothermal process what about ΔU?
    2.what is difference b/w ΔU in isothemall process and not
    3.is there is any difference b/w intermolecular force of water at 10C and 0C
    4.in phase change does the molecule get kinetic energy(i think not)
     
    Last edited: Feb 7, 2016
  18. Feb 7, 2016 #17
    There is a ΔU for a phase change without temperature changing.
    In addition the any ΔU associated with any phase change or chemical reaction (at constant temperature), there can be sensible heat contributions, and also contributions from the effect of pressure on internal energy
    I'm a continuum mechanics guy, so I don't pay a whole lot of attention to what is happening on the molecular scale
     
  19. Feb 8, 2016 #18
    you said ΔU for a reaction is caused by the bond dissociation and formation.
    i want to know more about it(why here T remains same)
    why in phase change also T remains constant(if there isn't any kinetic energy change if solid/liquid→gas and vice-versa that cause T difference)
    then what cause ΔU
     
  20. Mar 7, 2016 #19
     
  21. Mar 7, 2016 #20
    I have no idea what this question means. Is the exothermic reaction carried out adiabatically or isothermally?
     
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